在这个问题中,给我们一个数字N。我们的任务是创建一个程序,以查找C ++中相邻元素之差的最大和。
我们将找到所有置换数组的相邻元素之间的绝对差的最大和。
让我们举个例子来了解这个问题,
N = 4
输出结果
7
All permutations of size 4 are : {1, 2, 3, 4} = 1 + 1 + 1 = 3 {1, 2, 4, 3} = 1 + 2 + 1 = 4 {1, 3, 2, 4} = 2 + 1 + 2 = 5 {1, 3, 4, 2} = 2 + 1 + 2 = 5 {1, 4, 2, 3} = 3 + 2 + 1 = 6 {1, 4, 3, 2} = 3 + 1 + 1 = 5 {2, 1, 3, 4} = 1 + 2 + 1 = 4 {2, 1, 4, 3} = 1 + 3 + 1 = 5 {2, 3, 1, 4} = 1 + 2 + 3 = 6 {2, 3, 4, 1} = 1 + 1 + 3 = 5 {2, 4, 1, 3} = 2 + 3 + 2 = 7 {2, 4, 3, 1} = 2 + 1 + 2 = 5 {3, 1, 2, 4} = 2 + 1 + 2 = 5 {3, 1, 4, 2} = 2 + 3 + 2 = 7 {3, 2, 1, 4} = 1 + 1 + 3 = 5 {3, 2, 4, 1} = 1 + 2 + 3 = 6 {3, 4, 1, 2} = 1 + 3 + 1 = 5 {3, 4, 2, 1} = 1 + 2 + 1 = 4 {4, 1, 2, 3} = 3 + 1 + 1 = 5 {4, 1, 3, 2} = 3 + 2 + 1 = 6 {4, 2, 1, 3} = 2 + 1 + 2 = 5 {4, 2, 3, 1} = 2 + 1 + 2 = 5 {4, 3, 1, 2} = 1 + 2 + 1 = 4 {4, 3, 2, 1} = 1 + 1 + 1 = 3
为了解决这类问题,我们需要找到置换的总和。
这里是N的不同值的相邻元素的最大差之和的一些值。
N = 2, maxSum = 1 N = 3, maxSum = 3 N = 4, maxSum = 7 N = 5, maxSum = 11 N = 6, maxSum = 17 N = 7, maxSum = 23 N = 8, maxSum = 31
这个和看起来像是N的加法+ N个词的和
maxSum = S(N)+ F(N)S(N)= n(n-1)/ 2,而F(N)是N的未知函数
让我们使用S(N),maxSum(N)找到F(N)。
F(2) = 0 F(3) = 0 F(4) = 1 F(5) = 1 F(6) = 2 F(7) = 2 F(8) = 3
从这里,我们可以得出F(N)是Int(N / 2-1)。F(N)每N的第二个值增加一次,最初对于2和3则为零。
这样就可以得出maxSum的公式,
maxSum = N(N-1)/2 + N/2 - 1 maxSum = N(N-1)/2 + N/2 - 2/2 maxSum = ( N(N-1) + N - 2 )/2 maxSum = ( (N^2) - N + N - 2 )/2 maxSum = ((N^2) - 2 )/2
使用此公式,我们可以找到任何给定N值的maxSum值。
该程序说明了我们解决方案的工作原理,
#include <iostream> using namespace std; int calcMaxSumofDiff(int N){ int maxSum = 0; maxSum = ((N*N) - 2) /2 ; return maxSum; } int main(){ int N = 13; cout<<"The maximum sum of difference of adjacent elements is "<<calcMaxSumofDiff(N); return 0; }
输出结果
The maximum sum of difference of adjacent elements is 83